1. (12.1)(condi0onal(probability( 2. (12.2)(independence( · example(12.1((tay]sachs(disease)(•...
TRANSCRIPT
Chapter 12: Condi0onal Probability & Independence
1. (12.1) Condi0onal Probability 2. (12.2) Independence
Mo0va0ng Example
• In this chapter we explore finding the probability of an event given certain condi0ons or prior informa0on.
• For example, consider the experiment of rolling two dice. The probability of the event A = “sum of 6” is easy to find:
• Suppose, however, that you roll two dice and you don’t look at them. I tell you that you rolled doubles. Now what is the probability that the sum is 6?
• Solu0on: Let event A = “sum of 6” and B = “rolled doubles.” We now want to know the probability of event A given event B. Intui0vely, we expect that the answer should be 1/6 since there are six ways to roll doubles and only one of them sums to 6.
1. (12.1) Condi0onal Probability
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P A( ) =AS
=15, 24, 33, 42, 51{ }
S=536
Defini0on: Condi0onal Probability
• Let B be an event with P(B) > 0. Then we define the condi&onal probability of event A given B by:
• Heuris0cally, consider the diagram below and adopt the “dartboard” point of view described in Chapter 12:
1. (12.1) Condi0onal Probability
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P AB( ) =P A∩ B( )P B( )
Given that event B has occurred, we know that our “dart” has landed somewhere in B. Thus, the probability that it also landed in A is the area of the shaded region, R, divided by the area of rectangle B.
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=area of R( ) area of S( )area of B( ) area of S( )
=area of Rarea of B
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R
Mo0va0ng Example (Again)
• Recall our solu0on for the mo0va0ng example: Let event A = “sum of 6” and B = “rolled doubles.” Intui0vely, we expect that the the probability of A given B should be 1/6 since there are six ways to roll doubles and only one of them sums to 6.
• We now check this against our defini0on:
1. (12.1) Condi0onal Probability
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P AB( ) =P A∩ B( )P B( )
=33{ } S
11,22,33,44,55,66{ } S=1 366 36
=16
Example 12.1 (Tay-‐Sachs Disease)
• Tay-‐Sachs disease is a serious disorder of the nervous system that usually results in death by age 2 or 3. Affected individuals have genotype `, while normal (non-‐affected) individuals have genotype Tt or TT.
• Judy has a li`le brother with Tay-‐Sachs disease and is worried she may carry the recessive allele. What is the probability of this?
• Solu0on: First, no0ce that both of Judy’s parents must be Tt if her li`le brother has the disease. (We assume that neither parent can be ` since they each lived to adulthood.)
1. (12.1) Condi0onal Probability
Example 12.1 (Tay-‐Sachs Disease)
• The Punne` square gives the possible genotypes for Judy without being given any other informa0on: Since Judy does not have the disease, we
eliminate the ` as a possibility. Looking at the Punne` square, we intui0vely guess
that the probability should be 2/3 . Let us show this is true using the defini0on of condi0onal probability.
• Let A = “Judy is Tt” and B = “Judy is not `. Then,
1. (12.1) Condi0onal Probability
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P AB( ) =P A∩ B( )P B( )
=Tt,tT{ } S
TT,Tt,tT{ } S=2 43 4
=23
Example 12.2 (Drug Tes0ng)
• A test for a new sleeping pill involved 200 individuals where 100 of the individuals were given the sleeping pill, and the other 100 were given a sugar pill. The results of the test are shown in the following table: What is the probability that if
you take the sleeping pill you will sleep be`er?
• Solu0on: Let event A = “took sleeping pill,” and event B = “slept be`er.” We want to find the probability that you will sleep be`er given that you took the sleeping pill, i.e. P(B|A).
1. (12.1) Condi0onal Probability
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P BA( ) =P B∩ A( )P A( )
=B∩ A SA S
=71 200100 200
= 0.71
Example 12.2 (Drug Tes0ng)
What is the probability that if you slept be`er, you took the sleeping pill?
• Solu0on: (This is the converse of the previous ques0on.) Let event A = “took sleeping pill,” and event B = “slept be`er.” Now we want to find P(A|B):
1. (12.1) Condi0onal Probability
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P AB( ) =P A∩ B( )P B( )
=A∩ B SB S
=71 200129 200
= 0.55
Some More Probability Laws: P(Ā|B)
• Let A & B be events in S. Then,
• The deriva0on is straigheorward:
1. (12.1) Condi0onal Probability
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P A B( ) =1− P A B( )
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P A B( ) =P A ∩ B( )
P B( )=
P B∩ A ( )P B( )
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=P B( ) − P B∩ A( )
P B( )
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=1−P A∩ B( )P B( )
=1− P AB( )
Note, however, that this does not work for the other posi0on. That is:
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P A B ( ) ≠1− P A B( )
Probability Law: P(A∩B)
• Let A & B be events in S. If If P(B) ≠ 0 then,
• The deriva0on follows immediately from the defini0on of
condi0onal probability:
• No0ce that since set intersec0on is commuta0ve (that is, A∩B = B∩A), we automa0cally get:
1. (12.1) Condi0onal Probability
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P A∩ B( ) = P AB( )P B( )
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P AB( ) =P A∩ B( )P B( )
⇒ P A∩ B( ) = P AB( )P B( )
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P A∩ B( ) = P B A( )P A( )
Example 12.3.a (Beetle Sampling without Replacement)
We have a popula0on of 150 beetles. Thirty percent have wings and the rest are wingless. You select a beetle at random and record whether or not it has wings, and do not put it back. Then you select another beetle.
What is the probability that the first and second beetle have wings?
Solu0on: Let event A = “first beetle is winged,” and event B = “second beetle is winged.” First we note that, at the start, there are 0.3×150 = 45 winged beetles. Then:
1. (12.1) Condi0onal Probability
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P B∩ A( ) = P B A( )P A( )
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=44149"
# $
%
& ' ⋅
45150"
# $
%
& ' ≈ 0.0886
Defini0on
Suppose the probability of some event is not affected by the occurrence of another event. It seems natural in such a case to consider these events independent of one another. We formalize the no0on: We say events A and B are independent if P(A|B) = P(A).
No0ce that, formally, if we swap the le`ers A & B in the defini0on, we get P(B|A) = P(B). We should verify that this is consistent with our defini0on for condi0onal probability: Suppose that A and B are independent. Then,
2. (12.2) Independence
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P A( ) = P AB( )
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=P A∩ B( )P B( )
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P A( )P B( ) = P A∩ B( )
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⇒
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⇒
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P B( ) =P B∩ A( )P A( )
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= P B A( )We want to take special note of this rela0onship.
Probability Law
• Let A & B be independent events in S. Then,
• No0ce, when two events are mutually exclusive, the
probability of either/or event is the sum of the probabili0es of each event. However, when two events are independent, the probability of both events occurring is the product of the probabili0es of each event:
2. (12.2) Independence
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P A∩ B( ) = P A( )P B( )
P(A∪B) P(A∩B)
Mutually exclusive P(A) + P(B)
Independent ? P(A) P(B)
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∅
Mutually Exclusive & Independent?
• Recall that if events A and B are mutually exclusive, then P(A∩B) = 0. So if P(B) ≠ 0, then by the defini0on of condi0onal probability we have:
• Now we ask, can events A and B be both independent and
mutually exclusive? • Suppose, they were then P(A|B) = P(A) and P(A|B) = 0, and
thus P(A) = 0. Furthermore, P(B|A) = P(B) and P(B|A) = 0, and thus P(B) = 0.
• Therefore, if events A and B are both independent and mutually exclusive, then P(A) = 0 and P(B) = 0. That is, events that are both mutually exclusive and independent are only events with zero probability.
2. (12.2) Independence
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P AB( ) =P A∩ B( )P B( )
= 0
Example 12.4
Two dice are tossed, one at a 0me. Let A = “6 on first die”, B = “sum of 7”, and C = “sum of 8”. Which events are independent? Which events are mutually exclusive?
Solu0on: B&C: Clearly, B and C are mutually exclusive with non-‐zero
probabili0es, and therefore not independent. To answer the ques0on for A&B and A&C, we need the following
probabili0es: There are six possible outcomes on a die and only one way to roll a 6. Thus, P(A) = 1/6 . In rolling two dice, there are 6 × 6 = 36 possible outcomes. There are six ways to get a “sum of 7”, {(16), (25), (34), (43), (52), (61)}. Thus, P(B) = 6/36 = 1/6. There are five ways to get a “sum of 8”, {(26), (35), (44), (53), (62)}. Thus, P(C) = 5/36. Event (A∩B) can only occur with a 6 on the first die and a 1 on the second die. Thus, P(A∩B) = 1/36. Event (A∩C) can only occur with a 6 on the first die and a 2 on the second die. Thus, P(A∩C ) = 1/36.
2. (12.2) Independence
Example 12.4
And we have the following: A&B:
Thus, A&B are independent (and not mutually exclusive.) A&C:
Thus, A&C are not independent (and not mutually exclusive.)
2. (12.2) Independence
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P AB( ) =P A∩ B( )P B( )
=1 366 36
=16
= P A( )
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P AC( ) =P A∩C( )P C( )
=1 365 36
=15≠ P A( )
When to Assume Independence?
When is it reasonable to assume independence? Some examples relevant to our chapter:
1. Result of second sample is independent of first if first is placed back into sampling pool.
2. Sex of second child is independent of first child. 3. Random selec0on of alleles on genes located on separate
chromosomes (Mendel’s Second Law of Independent Assortment).
4. The genotypes of non-‐related individuals are independent events.
2. (12.2) Independence
Example 12.5 (Beetle Sampling with Replacement)
Suppose we have a popula0on of 150 beetles. Forty-‐five of the beetles have wings and the rest are wingless. You select a beetle at random and record whether or not it has wings, and then put it back amongst the other beetles. Then you select another beetle. Let events A = “first beetle has wings” and B = “second beetle has wings”
What is the probability that (a) the first and second beetle have wings? and (b) the first beetle has wings and the second beetle does not have wings?
Solu0on: (a) (b)
2. (12.2) Independence
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P A∩ B( ) = P A( )P B( ) = 0.3× 0.3 = 0.09
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P A∩ B ( ) = P A( )P B ( ) = 0.3× 0.7 = 0.21
Example 12.6 (Blood Typing)
Suppose Jacob has O+ blood with genotype OO/Rr, and Anna has B− blood with genotype BO/rr. What is the probability that a child of Jacob and Anna will have (a) B+ blood, and (b) O− blood?
Solu0on: We start by making a Punne` square for each gene: Thus: (a) (b)
2. (12.2) Independence
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P B +( ) = P B( )P +( ) = 0.5 × 0.5 = 0.25
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P O−( ) = P O( )P −( ) = 0.5 × 0.5 = 0.25
Example 12.7 (Tay Sachs Disease)
This example uses both condi0onal probability and independence.
Jack and Judy each had brothers afflicted with Tay-‐Sachs disease. From Example 12.1, the probability of being a carrier if your sibling has Tay-‐Sachs disease is 2/3. What is the probability that Jack and Judy have a child with Tay-‐Sachs disease?
Solu0on: The only way the child can be ` is if Jack and Judy are both Tt. Let events A = “Judy is Tt”, B = “Jack is Tt”, C = “child is `”. Thus:
2. (12.2) Independence
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P C∩ A∩ B( )( ) = P C A∩ B( )( )P A∩ B( )
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= P C A∩ B( )( )P A( )P B( )
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=14⋅23⋅23
=19
Example 12.8 (Family Planning)
Assume the births of boys and girls are equiprobable and independent. Let events A = “first child is a girl,” B = “second child is a girl,” and C = “third child is a girl.” These are all independent events.
(a) What is the probability that in a family with two children, the children are both girls?
(b) What is the probability that in a family with three children,
the children are all girls?
2. (12.2) Independence
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P A∩ B( ) = P A( )P B( ) =12⋅12
=14
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P A∩ B∩C( ) = P A( )P B( )P C( ) =12⋅12⋅12
=18
Example 12.8 (Family Planning)
(c) What is the probability that in a family with six children, the children are all the same sex? The probability that all six children are the same sex equals the probability they are all boys plus the probability they are all girls. Let events A1, A2, . . . , A6 be the events that the first, second, . . . , sixth (respec0vely) child is a girl. Then:
2. (12.2) Independence
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P A1∩ A2∩∩ A6( ) = P A1( )P A2( )P A6( ) =12#
$ % &
' ( 6
=164
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P A 1∩ A 2∩∩ A 6( ) = P A 1( )P A 2( )P A 6( ) =12#
$ % &
' ( 6
=164
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P all same sex( ) =1
64+
164
=132
Example 12.8 (Family Planning)
(d) What is the probability that in a family with six children, there is at least one girl? Let A = “at least one girl”. Then Ā = “no girls”, i.e. all boys. Then:
2. (12.2) Independence
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P A( ) =1− P A ( ) =1− 164
=6364
≈ 0.9844
Homework
Chapter 12: 1ab, 2, 4, 5, 6, 7, 9 Some answers: